1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Implicit multivariable derivative of a spherical cap

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a spherical cap, for which the surface area and volume is

    [tex]A(a,h) = \pi(a^2 +h^2)[/tex]
    [tex]V(a,h) = \frac{\pi h}{6}(3a^2 +h^2)[/tex]

    What would the aspect ratio [itex]dA/dV[/itex] be?

    3. The attempt at a solution

    Clearly we would have

    [tex]dA = 2\pi a da + 2\pi h dh[/tex]
    [tex]dV = \pi ha da + \frac{\pi}{2}(a^2+h^2) dh [/tex]

    but what would the chain rule look like? Surely, it should have more terms than just

    [tex]\frac{dA}{dV } = \frac{\partial A}{\partial a} \frac{\partial a}{\partial V} + \frac{\partial A}{\partial h}\frac{\partial h}{\partial V}[/tex]

    , right?
     
  2. jcsd
  3. Jun 9, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why do you think it should have more terms? a and h are the only variables.
     
  4. Jun 9, 2014 #3
    If [itex]dV, dA[/itex] were just functions of one variable, it would be straightforward. But here there are two variables, and perhaps I would have anticipated a chain rule of the form

    [tex]\frac{dA}{dV} = \frac{\partial A}{\partial a}\left( \frac{\partial a}{\partial V} + \frac{\partial h}{\partial V} \right) + \frac{\partial V}{\partial a}\left( \frac{\partial a}{\partial A} + \frac{\partial h}{\partial A} \right) [/tex]

    but I'm just guessing.
     
  5. Jun 9, 2014 #4

    pasmith

    User Avatar
    Homework Helper

    On dimensional grounds, I would expect [tex]
    A = V^{2/3} f\left(\frac ah\right)[/tex] for some function [itex]f[/itex] which can be determined from the formulae for [itex]A[/itex] and [itex]V[/itex]. Thus [itex]A[/itex] is a function of both [itex]V[/itex] and [itex]a/h[/itex].
     
  6. Jun 9, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The second part has V "in the numerator" that can't be right! "On dimensional grounds", as pasmith puts it, V has units of "length cubed", A has units of "length squared, and h and a have units of "length". [itex]\partial V/\partial a[/itex] has units of "length squared", and [itex]\partial a/\partial A[/itex] and [itex]\partial h/\partial A[/itex] have units of "1 over length" so the second term has units of length. But [itex]dA/dV[/itex] has units of "1 over length".
     
  7. Jun 11, 2014 #6
    Thanks for the advice HallsofIvy and pasmith!

    Pasmith, yes, you were right, I was able to evaluate the derivative by expressing the functions in terms of [itex]f(h/a)[/itex]. Thanks for the insight!
     
  8. Dec 19, 2014 #7
    Dear Pasmith, I have been looking at this problem again and in fact I don't think it is possible to find the derivative [itex]dA/dV[/itex] like you mentioned.

    Let [itex]x\equiv a/h.[/itex]
    [itex]A^3 = V^2 f(a/h) = V^2 f(x)[/itex]
    [itex]3A^2 dA = 2V dV f(x) + V^2 \frac{df}{dx} dx[/itex]

    The complication comes in when you are trying to solve for [itex]dA/dV[/itex] and you get

    [itex]\frac{dA}{dV} = ... + \frac{V^2}{3A^2} \frac{df}{dx} \frac{dx}{dV}.[/itex]

    But note that [itex]V=\frac{\pi h}{6}(3a^2+h^2)[/itex]. Since the powers of [itex]h, a[/itex] are different, there is no way to evaluate the derivative [itex]dx/dV[/itex] without leaving behind some residual terms of [itex]a, h[/itex]. The goal in the first place was to evaluate the derivative solely in [itex]x\equiv a/h[/itex]. Therefore we can't parameterise the equation like you said here.

    Can you help, Pasmith, or anybody? This problem has been bothering me for a while.
     
  9. Dec 19, 2014 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A simpler approach is to use the relationship between ##a## and ##h## to express ##A## and ##V## in terms of ##h## and the radius of the sphere. That reduces it to a single variable problem.
     
  10. Dec 19, 2014 #9
    [itex]a , h[/itex] are also related by a contact angle [itex]\theta[/itex]. But you can't express a spherical cap uniquely in just that one variable.

    I guess I don't really understand what you meant. From what I understand a spherical cap is uniquely described by at least two parameters - one possible combination is the the base width and height [itex]a,h[/itex], and another combination is the radius of curvature and angle [itex]r,\theta[/itex].
     
  11. Dec 19, 2014 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I guess from what you said that this is not a spherical cap on a given sphere?
     
  12. Dec 19, 2014 #11
    Yes, that's right, the cap is not on a given sphere. There are two parameters.
     
  13. Dec 19, 2014 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So, are you not trying to calculate ##\frac{\partial A}{\partial V}## ?
     
  14. Dec 19, 2014 #13
    That is what I am calculating. As we know, for a sphere, [itex]V(r), A(r)[/itex] and it's easy to calculate [itex]\frac{dA}{dV} = \frac{dA}{dr} \frac{dr}{dV}[/itex]. But what happens when these two quantities are functions of more than one variable, ie, [itex]V(a,h), A(a,h)[/itex]? Then how would we calculate [itex]\partial A/\partial V[/itex]?
     
  15. Dec 19, 2014 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think your problem is that you also need to specify which variable is constant for your partial derivative. For example, you could express A as a function of V and h, say:

    ##A = f(V, h)##

    And then you could calculate [itex]\partial A/\partial V = \partial f/\partial V [/itex] where h is held constant.

    But, you could also find:

    ##A = g(V, a)##

    And then you could calculate [itex]\partial A/\partial V= \partial g/\partial V [/itex] where a is held constant. And these would be different functions with different numerical values.

    Or, you could hold ##a/h## constant etc.

    That's why I assumed it was a fixed sphere, when you have V as a function of A with no other independent variables and can find ##\frac{dA}{dV}##
     
  16. Dec 19, 2014 #15

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your question is not well-defined (as many others have also indicated). However, it is not very difficult to extract answers to more pointed questions. We have
    [tex] \Delta A = 2 \pi a \, \Delta a + 2 \pi h \, \Delta h, \\
    \Delta V = \pi h a \, \Delta a + \frac{\pi}{2}(a^2+h^2) \, \Delta h \\
    \text{so}\\
    \frac{\Delta A}{\Delta V} =
    \frac{2 \pi a \, \Delta a + 2 \pi h \, \Delta h}{\pi h a \, \Delta a + (\pi /2)(a^2+h^2) \, \Delta h}
    [/tex]
    This implies the following.
    (1) If ##a## is a function of ##h## we get
    [tex] \frac{dA}{dV} = \frac{2 \pi a (da/dh) + 2 \pi h}{\pi h a (da/dh) + (\pi/2) (a^2 + h^2)}[/tex]
    (2) If ##h## is a function of ##a## we get
    [tex] \frac{dA}{dV} = \frac{2 \pi a+ 2 \pi h (dh/da)}{\pi h a + (\pi/2)(a^2+h^2) (dh/da)} [/tex]
    (3) If ##h## is a constant, the derivative ##(dA/dV)_h## --which is Thermodynamics-style notation, indicating that ##h## is held constant--has the form
    [tex] \left( \frac{d A}{d V} \right)_{h} = \frac{ 2 \pi a}{\pi h a}.[/tex]
    (4) If ##a## is a constant, the derivative ##(dA/dV)_a## is
    [tex] \left( \frac{dA}{dV}\right)_a = \frac{2 \pi h}{(\pi/2)(a^2+h^2)}[/tex]

    You get many different answers, depending on what you mean by the question to begin with.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Implicit multivariable derivative of a spherical cap
  1. Vol of spherical cap? (Replies: 4)

  2. Implicit Derivation (Replies: 9)

Loading...